A sheet of paper 90 cm-by-66 cm is made into an open box (i.e. there's no top), by cutting x-cm squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box. Give your answer in the simplified radical form. x =?? is the max.

Respuesta :

Answer:

[tex]26 - \sqrt{181}[/tex] cm

Step-by-step explanation:

The volume of the box is:

V = height * length * width

V = x*(66 - 2*x)*(90 - 2*x)

V = (66*x - 2*x^2)*(90 - 2*x)

V = 5940*x - 132*x^2 - 180*x^2 + 4*x^3

V = 4*x^3 - 312*x^2 + 5940*x

where x is the length of the sides of the squares,  in cm.

The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

subject to:

x > 0

2*x < 66 <=> x < 33

In the maximum, the first derivative of V, dV/dx, is equal to zero

dV/dx = 12*x^2 - 624*x + 5940

From quadratic formula

[tex]x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)} [/tex]

[tex]x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)} [/tex]

[tex]x = \frac{624 \pm \sqrt{104256}}{24} [/tex]

[tex]x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24} [/tex]

[tex]x = \frac{624 \pm 8*3*\sqrt{181}}{24} [/tex]

[tex]x_1 = \frac{624 + 24*\sqrt{181}}{24} [/tex]

[tex]x_1 = 26 + \sqrt{181}[/tex]

[tex]x_2 = \frac{624 - 24*\sqrt{181}}{24} [/tex]

[tex]x_2 = 26 - \sqrt{181}[/tex]

But [tex]x_1 > 33[/tex], then is not the correct answer.

The value of x that maximizes the volume of the box is:

[tex]\mathbf{x = 26 - \sqrt{181}}[/tex]

The dimension of the paper is given as:

[tex]\mathbf{Length = 90}[/tex]

[tex]\mathbf{Width = 66}[/tex]

When the cut-out (x) is removed, we have:

[tex]\mathbf{Length = 90 - 2x}[/tex]

[tex]\mathbf{Width = 66 - 2x}[/tex]

[tex]\mathbf{Height =x}[/tex]

So, the volume is:

[tex]\mathbf{Volume = Length \times Width \times Height}[/tex]

This gives

[tex]\mathbf{V(x)= (90 - 2x) \times (66 - 2x )\times x}[/tex]

Expand

[tex]\mathbf{V(x)= 5940x -312x^2 + 4x^3}[/tex]

Differentiate

[tex]\mathbf{V'(x)= 5940 -624x + 12x^2}[/tex]

Set to 0

[tex]\mathbf{5940 -624x + 12x^2 = 0}[/tex]

Divide through by 12

[tex]\mathbf{495 -52x + x^2 = 0}[/tex]

Rewrite as:

[tex]\mathbf{ x^2-52x + 495= 0}[/tex]

Using a calculator, the possible values of x are:

[tex]\mathbf{x = 26 \pm \sqrt{181}}[/tex]

The value [tex]\mathbf{x = 26+ \sqrt{181}}[/tex], is greater than the dimension of the box.

Hence, the value of x that maximizes the volume of the box is:

[tex]\mathbf{x = 26 - \sqrt{181}}[/tex]

Read more about volumes at:

https://brainly.com/question/1578538

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